What does $d_{\textbf{a}} f$ mean? I have a question regarding the differential $d_{\textbf a} f$. 
Suppose we have the function $f(x,y)= xy$, and the vectors $\textbf a = (1,1)$ and $\textbf u = (2,1)$. Then, if I understand this correctly,
$$d_{\textbf a} f(\textbf u) = \nabla f(\textbf a) \cdot \textbf u = (1,1)\cdot (2,1) = 2+1 = 3,$$
where $\nabla f(\textbf a) = (\partial f/\partial x, \partial f/\partial y)$. But what if my assignment is to calculate $d_{\textbf a} f$? I don't know what it means. Do they want me to calculate $d_{\textbf a} f(x,y) = (1,1)\cdot (x,y) = x+y$, or something else?
Edit: Note that it is not the directional derivative that I'm asking about.
 A: Essentially, you have worked out everything already, but there seems to be a bit of confusion about the definitions, so let me try to set this straight.
The differential of $f$ at the point $\mathbf{a} \in \mathbb{R}^2$ is the row matrix
$$ d_{\mathbf{a}}f = \begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) & \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix}.$$
Now if you write $d_{\mathbf{a}}f (\mathbf{u})$ for $\mathbf{u} = \begin{pmatrix} u_1 \\\ u_2 \end{pmatrix} \in \mathbb{R}^2$ you're meaning the matrix product
$$d_{\mathbf{a}}f (\mathbf{u}) = 
\begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) & \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix} \cdot \begin{pmatrix} u_1 \\\ u_2 \end{pmatrix} = \frac{\partial}{\partial x} f(\mathbf{a}) \cdot u_1 + \frac{\partial}{\partial y}f (\mathbf{a}) \cdot u_2 .$$
On the other hand, $\nabla f (\mathbf{a})$ is the column vector
$$
\nabla f (\mathbf{a}) = 
\begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) \\\ \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix}$$
and when you're writing $\nabla f (\mathbf{a}) \cdot \mathbf{u}$ you're meaning the scalar product
$$\nabla f( \mathbf{a}) \cdot u = \begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) \\\ \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix} \cdot  \begin{pmatrix} u_1 \\\ u_2 \end{pmatrix} = \frac{\partial}{\partial x} f(\mathbf{a}) \cdot u_1 + \frac{\partial}{\partial y}f (\mathbf{a}) \cdot u_2 .
$$
So we see that for $f(x,y) = xy$ 
$$d_{\mathbf{a}}f = \begin{pmatrix} y & x \end{pmatrix} \qquad \text{while} \qquad
\nabla f (\mathbf{a}) = \begin{pmatrix} y \\\ x \end{pmatrix}.$$
Now the confused reaction was due to the fact that the notation used here for the derivative of $f$ at the point $\mathbf{a}$ is often used as the directional derivative, and as you rightly pointed out in a comment, we have the relations
$$ D_{\mathbf{u}} f (\mathbf{a}) : = d_{\mathbf{a}} f (\mathbf{u}) = \nabla f(\mathbf{a}) \cdot \mathbf{u},$$
and everything should be fine now, no?
Since you made the computations yourself already, I'll not repeat them here.
A: In the case of functions $f\colon \mathbb{R}^n \to \mathbb{R}$, like $f(x,y) = xy$ as you have, the differential $d_af$ is the same thing as the gradient $\nabla f(a)$.
